a+b) 2Zn+O2--->2ZnO
x------1/2x--------x
4Al+3O2---->2Al2O3
y--------0,75y----0,5y
Ta có
n O2=8,96/22,4=0,4(mol)
Ta có hệ pt
\(\left\{{}\begin{matrix}65x+27y=23,8\\0,5x+0,75y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,4\end{matrix}\right.\)
%m Zn=0,2.65/23,8.100%=55%
%m Al=100-55=45%
c)m ZnO=0,2.81=16,2(g)
m Al2O3=102.0,2=20,4(g)
%m ZnO=\(\frac{16,2}{16,2+10,4}.100\%=44\%\)
%m Al2O3=100-44=64%